Chapter 4, Problem 88A

(a)

To determine

To Calculate: The acceleration of the block.

Answer to Problem 88A

  $3{\text{m/s}}^2$

Explanation of Solution

Given:

Mass of block1 $m_1=2\text{kg}$

Mass of block2, $m_2=4\text{kg}$

Mass of block3, $m_3=6\text{kg}$

Applied force, $F=36\text{N}$

Formula used:

Newton’s law of motion:

  $F=ma$

Where,

  $a$ is the acceleration.

  $m$ is the total mass

Calculation:

Total mass of the system,

  $\begin{array}{l}m=m_1+m_2+m_3\\ m=2+4+6\\ m=12\text{kg}\end{array}$

From the newton’s second law of motion

Force can be given as:

  $\begin{array}{l}F=ma\\ a=\frac{F}{m}\\ a=\frac{36}{12}\\ a=3{\text{m/s}}^2\end{array}$

Conclusion:

the acceleration of the block is $a=3{\text{m/s}}^2$

(b)

To determine

To Calculate: The tension forces in each string.

Answer to Problem 88A

  $F_{T_1}=6\text{N}$

  $F_{T_2}=18\text{N}$

Explanation of Solution

Given:

Mass of block1 $m_1=2\text{kg}$

Mass of block2, $m_2=4\text{kg}$

Mass of block3, $m_3=6\text{kg}$

Applied force, $F=36\text{N}$

Formula used:

Newton’s law of motion:

  $F=ma$

Where,

  $a$ is the acceleration.

  $m$ is the total mass

Calculation:

  

Net force on the block3 is towards the right due to applied force and tension is on the left.

Therefore,

  $m_{3}a=F-F_{T_2}\mathrm{.....}\left(1\right)$

  

Net force on the block 2 is right side due to tension $F_{T_2}$ and $F_{T_1}$ is the tension in left side.

Therefore,

  $m_{2}a=F_{T_2}-F_{T_1}\mathrm{.....}\left(2\right)$

  

And the net force on the block 1 is in right side due to tension $F_{T_1}$

Therefore, $m_{1}a=F_{T_1}\mathrm{.....}\left(3\right)$

Plugging the values in equation (3)

  $\begin{array}{l}2\times 3=F_{T_1}\\ F_{T_1}=6\text{N}\end{array}$

From equation (1)

  $\begin{array}{l}6\times 3=36-F_{T_2}\\ F_{T_2}=18\text{N}\end{array}$

Conclusion:

Tension in the string is $F_{T_1}=6\text{N}$ and $F_{T_2}=18\text{N}$ .